I would simply identify your bilinear map and the induced map on the tensor product, and simply say they are the same thing. The universal product of the tensor product tells us that I would not run into much trouble.
–
Mariano Suárez-Alvarez♦Aug 28 '11 at 17:57

5 Answers
5

The correspondence you describe is part of the definition of the tensor product: $V \otimes W$ is defined to have the universal property that for any $U$, we have $$\operatorname{Hom}(V \otimes W, U) = \operatorname{Bil}(V \times W, U).$$ I wouldn't even give it a different name: the bilinear form is the same as the map out of the tensor product.

I don't know a specific name for that, but I would call it associated. I wouldn't call it induced, because the map $Bil(V) \to (V \otimes V)^\*$ is one-to-one, since every linear $T \in (V\otimes V)^\*$ induces a bilinear form on $V$ by sending $(v,w) \mapsto T(v\otimes w)$ and this clearly is the inverse.

And symmetry of bilinear forms can be encoded in the tensor product: if we define the symmetric tensor product $V\otimes_s V$ to be the subspace of $V \otimes V$ spanned by the "symmetric tensors" $v\otimes_s w = (v\otimes w + w\otimes v)/2$, then there is a canonical bijection between linear forms on $V\otimes_s V$ and symmetric bilinear forms on $V\times V$.

There may be some interest in a variant: the natural map $V\otimes V^{\star}$ to $End(V)$ hits all finite-rank endomorphisms of $V$, and the bilinear map $V\times V^\star\to k$ "induces" trace on finite-rank endos. Thus, some name like "trace" is quite nearby to the literal question.